Engine Masters. Daily report.

[MadBill]

Of course there's the issue of standing Vs. rolling starts, possibly a traction limit to contend with in the lower gears and the desirability of not having to upshift mid-corner exit, but the starting point is to hit the limiter in high gear just before the braking point on the fastest straight, then juggle the intermediate ratios to provide a smaller RPM drop in each succeeding higher gear.

Just to make things interesting there's also sometimes a tradeoff to be made between a first gear that will get the car through a tortuous pit lane without smoking the clutch, Vs. one which will be useable on track. No wonder people like Bill Mitchell get big bucks for a gearing software program!

[A Atwood]

On a round track, where there are no shifts. The ideal gearing is one that puts the rpm of peak hp near mid way of the straight. The power level of the engine should be the same when coming off of the corner as it is just before letting off going into the turn. Basically the hump of the power curve is used on the straight. This puts the maximum usable axle tq to the dirt. Call it the highest average axle tq if you want (which it is), but it is the highest total axle tq that can be put to the ground.
Traction is a totally different animal.

ARN

[F1Fever]

The area under the torque curve from the torque peak to the power peak is most important. Not average anything else.

I just can't agree with that. at least not in reference to the atlantic car and the way you posted it appears as though that is what you are saying...

also, those graphs, are they from actual data or a sim program ?

[Stan Weiss]

For those who might be interested in the details related to the graphs that Stan was nice enough to post.

The car is an actual Toyota Formula Atlantic which was run at the Montreal Grand Prix (a road course). A four valve, four cylinder, NA Toyota engine, 1600cc making about 230 hp. It is rev limited to 10,000 rpm. Weight is 1340 lbs, frontal area is 11.61 sq ft., aero Cd is .62. Calculations include 15% driveline losses and 2% rolling resistance. Gear ratios, rear end ratio and tire size is indicated on the graph. The numbers on the curves are the rpms in 1000s.

David,
I am trying to run your numbers but do not come up with the same results. With no power loss my G's numbers are higher and with the 15% power loss my G's numbers are lower. does it look like I got any of the parameters wrong?
Stan

RPM = 10000
Tire Diameter = 23.0
Car Weight = 1340.0
Rear Gear Ratio = 4.11
T Gear1 = 2.50
T Gear2 = 1.93
T Gear3 = 1.59
T Gear4 = 1.32
T Gear5 = 1.05
T Gear6 = 0.0
T Gear7 = 0.0
Coefficient of Drag = 0.62
Frontal Area = 11.61
Percent Drive Train Power Loss = 0.0
Percent Rear End Power Loss = 15.0
Tire Rolling Radius = 11.5
Dyno Correction = 1.00

5200 92
5400 94
5600 99
5800 100
6000 101.5
6200 102.5
6400 110
6600 119.5
6800 122.5
7000 126
7200 131
7400 135
7600 138
7800 139
8000 140
8200 138
8400 136
8600 136
8800 136
9000 134.5
9200 131.5
9400 129
9600 127
9800 122
10000 119.5

Edited: Added Graph

[mbrooks]

my resources indicate that thrust must be calculated into the shift points.

nowhere is hp used in the calculation, torque is. thrust would be the x axis, mph would be the y axis.

a gear is run up to its maximum speed in rpm or the point at which thrust intersects the next ratio curve.

from race car engineering & mechanics by paul van valkenburgh, p.84

also a nice graph pictured if anybody can post it.

[F1Fever]

a gear is run up to its maximum speed in rpm or the point at which thrust intersects the next ratio curve.

If you don't mind, or anybody else, calculate it up how fast it is when you average the highest hp in each gear.. you can do this just like you said above, and double check that you will infact average the highest HP, shifting well past peak if you have a peaky motor with a wide spread between shifts.

The idea is to add you total HP from each point thruout your power curve up for a total and divide by the number of points. Take the following graph for example:


In it we have a peak Hp of 506 at 6250. if we shift at 6250 rpm and our rpm drops to 4500 rpm we have a total number of 460.1 HP {395 + 417 + 438 + 459 + 475 + 489 + 502 + 506= 3681/8= 460.125 hp average}... but if we wait until we have passed peak Hp, say 6750 we end up with 482.6 HP average {438 + 459 + 475 + 489 + 502 + 506 + 501 + 491= 3861/8= 482.625}. That is an increase of 22+ HP just by raising our shift points 500 rpm past peak.
please keep in mind this is not for every engine, a dyno sheet should be used along with the rpm spread that your transmission will produce at your selected shift points.

[Nick Campagna]

For mbrooks, this is what Stan and I said back on pg12 "Posted: Wed Nov 18, 2009 5:30 am "You have hit the nail on the head Bill. Thrust is what moves cars. Stan, your plotting of wheel torque vs wheel speed is also correct, as it shows where the shift points should be. The way we need to look at all this, is to realize that the shape of the torque curve after the peak determines the shape / location of the Hp curve. By shifting at Stan's crossover points, we maximize the available thrust to the car. I looked at this with a spreadsheet years ago, and have always been intrigued with others' arguments of Tq vs Hp. Hp can be looked at as a weighted function of Tq, with the weighting function being (rpm/5252). All that is needed now it to correct for the aerodynamic drag which is a power law function of speed to get the drag on the car for more accurate car performance predictions. The available thrust will be diminished by the available drag. Good work guys." and his graph showing intersecting curves of torque vs mph. Acceleration * mass = thrust, thus acceleration is linearly proportional to thrust or torque. The generalizations re 'how far past the hp peak' all assume the same shape of the decline of the post-peak torque curve, an obvious fallacy.

[gmrocket]

As an engine approaches the rev limit, the acceleration falls off rapidly, even if speed continues to increase. For this engine, closer gear steps would benefit since it is rev limited by the rules.

Now if the longest straight is not long enough to reach terminal velocity, what gearing changes should be made and how should it then be driven?

The more important point to be made is that a properly geared engine is neither driven at the torque peak nor at the power peak (at least not for very long). The area under the torque curve from the torque peak to the power peak is most important. Not average anything else.

For the EMC competition:
1. It would seem necessary to build an engine to work with the restraints of the dyno exhaust system.

2. Build an engine with high torque and small displacement since torque is counted twice in the scoring.

3. Obtain a sample of the actual fuel that will be used. A 93 octane pump gas has no real meaning whatsoever since there are many, many different 93 octane pump gasolines.

what do you mean by "torque is counted twice in the scoring"?

[F1Fever]

generalizations re 'how far past the hp peak' all assume the same shape of the decline of the post-peak torque curve, an obvious fallacy.

incorrect, read my post above yours, it is specific to power and gears.

if you want to fix up some graphs shifting to average the greatest HP as per my previous post it will become clear when overlayed with a graph that shifts at peak HP, or lower.

[rmcomprandy]

shifting at a higher rpm would accelerate faster b/c a higher average hp would be applied.

I would suggest that YOU get the use of an in car data aquision system with "G" meter capabilities in order to actually see how wrong you really are.

It is evident that you will not be convinced of anything other than your own oppinions until YOU actually do the testing yourself but, then you probably won't believe tha actual data either so, that might be useless too...

[F1Fever]

well there are apparently people here with software that will print out a graph for all of us...

[Stan Weiss]

OK I added to the torque and HP curves so we could shift at 10500 RPM et went from 11.5847 to 11.5665. Since I just picked a number it is possible that I have the torque falling off to quickly, but it is also possible that I do not have it falling off fast enough.

[A Atwood]

The best graph is the one on page 13 that plots G's vs. MPH. In that graph you can see that the best shifting would happen near 10,500 and drop near 8500. Now go and look at the hp graph (see above) and you will see why this is. It uses the most hp available. The ideal shift occurs EXACTLY when the horsepower has fallen to the point that the shift recovery rpm matches the hp before the shift. And this is also the SAME point that the rear wheel tq is the same. It is so simple. Just look at the hp curve and you can get a very good idea where the shift points should be without even looking at the tq graph.

ARN

[MadBill]

Another way to visualize this is to consider accelerating a vehicle with a continuously variable transmission (CVT) Surely everyone would agree that maintaining RPM at peak power would be the right approach? From Stan's graph of available torque in each gear, we can see that a CVT would create a curve that bridged the peaks of each discrete gear without falling into the valleys between them, thus increasing the overall average.

FWIW, many years ago I was doing tests on a ZR1 Corvette (with the Lotus 4-cam engine). I was moving the shift point higher and higher, looking for the magic point where the Gs after the shift matched the 'before'. The nominal power peak was as I recall, in the low six thousands, but I got right up to the rev limiter at 7,200 and still saw less accel after the shift. When I looked at the dyno sheet I knew why: Although the peak was indeed in the low sixes, the drop off was almost nil right out to the redline. For this engine and gear ratios, the optimum shift RPM would been ~ 7,600 and would have resulted in delivering an 'average power' output only a few horses less than the peak!

[mbrooks]

For mbrooks, this is what Stan and I said back on pg12 "Posted: Wed Nov 18, 2009 5:30 am "You have hit the nail on the head Bill. Thrust is what moves cars. Stan, your plotting of wheel torque vs wheel speed is also correct, as it shows where the shift points should be. The way we need to look at all this, is to realize that the shape of the torque curve after the peak determines the shape / location of the Hp curve. By shifting at Stan's crossover points, we maximize the available thrust to the car. I looked at this with a spreadsheet years ago, and have always been intrigued with others' arguments of Tq vs Hp. Hp can be looked at as a weighted function of Tq, with the weighting function being (rpm/5252). All that is needed now it to correct for the aerodynamic drag which is a power law function of speed to get the drag on the car for more accurate car performance predictions. The available thrust will be diminished by the available drag. Good work guys." and his graph showing intersecting curves of torque vs mph. Acceleration * mass = thrust, thus acceleration is linearly proportional to thrust or torque. The generalizations re 'how far past the hp peak' all assume the same shape of the decline of the post-peak torque curve, an obvious fallacy.

yeah,i saw this after my post, but i left it alone. interesting subject though. in addition:

if it was possible to have an infinite number of gears or an infinitely variable ratio, then the rpm could remain at peak tq. and thrust available curve would be a smooth curve. it should also be apparent that it is important to have a relatively flat engine tq. curve over as broad an rpm range as possible. if tq. falls off sharply or has noticeable dips due to intake ram tuning and exhaust tuning effects, it will be more difficult to fill the holes in the thrust curve by changing gear ratios.

it makes the common practive of rating race engines by peak hp. look relatively meaningless.

all the above from racecar engineering & mechanics by paul van valkenburgh

i think it just reiterates what some of you are showing with the graphs and previous posts is all.

oops, just saw bills post, almost a double.